Flux and luminosity

Builds on: Sections S3 and S4
Reading time: 25 min
Exercises: 3
Olympiad problems: 1

Above the air, each square metre facing the Sun receives about 1361 watts. At Mars, half as far again from the Sun, the same square metre receives only 586 watts, and at Jupiter only 50. The Sun gives out the same light in every case. What changes is how far that light has spread by the time it arrives. In this section we separate the two ideas, the power a star gives out and the power that reaches each square metre of a telescope, and find the simple rule that links them.

4.1.1Luminosity and flux

Before you readMake a guess first

Mars is about 1.5 times as far from the Sun as Earth is. How much sunlight does each square metre receive there, compared with at Earth?

The luminosity $L$ of a star is the total power it gives out as light, in watts. The Sun's luminosity is $3.828 \times 10^{26}$ W, and it is used as a unit, $L_{⊙}$ , for other stars. The flux $F$ is the power that crosses each square metre of a surface facing the star, in watts per square metre, and it is what a telescope measures.

A star sends its light out equally in all directions. At a distance $d$ that light is spread over a sphere of area $4 π d^{2}$ centred on the star, and since no light is lost on the way through empty space, the whole luminosity crosses that sphere. The flux is the luminosity shared out over the sphere.

ResultFlux from a star(4.1)

F = \frac{L}{4 π d^{2}}

$F$	the flux at distance $d$ , in W m⁻²
$L$	the luminosity of the star, in W
$d$	the distance from the star, in m

This is the inverse-square law. It assumes that the star shines equally in every direction and that nothing between the star and us absorbs its light. Section 4.4 deals with light lost on the way.

Example 4.1The Sun's luminosity from the solar constant

The flux of sunlight just above Earth's atmosphere, called the solar constant, is 1361 W m⁻². Earth is $1.496 \times 10^{11}$ m from the Sun. What is the Sun's luminosity?

Solution

Turning (4.1) round,

\begin{aligned} L & = 4 π d^{2} F \\ = 4 π \times (1.496 \times 10^{11})^{2} \times 1361 \\ = 3.83 \times 10^{26} W . \end{aligned}

AnswerThe Sun's luminosity is

3.83 \times 10^{26}

W, which agrees with the standard value of

3.828 \times 10^{26}

Quick checkSunlight at Jupiter

Jupiter is 5.20 au from the Sun. What is the flux of sunlight there, if it is 1361 W m⁻² at 1 au?

W m⁻²

4.1.2Comparing two stars

The inverse-square law makes comparisons easy, because $4 π$ cancels. Two stars of equal flux, one with luminosity $L_{1}$ at distance $d_{1}$ and one with $L_{2}$ at $d_{2}$ , satisfy $L_{1} / d_{1}^{2} = L_{2} / d_{2}^{2}$ , so a star 100 times as luminous looks just as bright from 10 times as far away. Astronomers use this all the time, since a distance follows at once from a luminosity known by other means and a measured flux.

Example 4.2A supernova as bright as the Sunfinish it

A supernova shines with $10^{10}$ times the luminosity of the Sun. How far from us would it have to be to look as bright as the Sun, in astronomical units?

The start

The supernova at distance $D$ and the Sun at 1 au give equal fluxes, so $10^{10} L_{⊙} / D^{2} = L_{⊙} / (1 au)^{2}$ .

Your stepNow find

D

Common errorsSection 4.1

Flux and luminosity are different quantities. Luminosity belongs to the star and does not depend on where we stand, while flux depends on our distance from it.
The flux falls as the square of the distance. A star twice as far away gives a quarter of the flux.
Distances must be in metres when the luminosity is in watts and the flux in W m⁻², or the units of the answer are wrong.

ExercisesSection 4.13 questions

4.1.1
Mars is 1.524 au from the Sun. What is the flux of sunlight there, if it is 1361 W m⁻² at 1 au?
W m⁻²
4.1.2
Two stars give the same flux at Earth, but the first is twice as far away as the second. How many times as luminous is the first?
times
4.1.3
A space probe moves from 1 au to 3 au from the Sun. What happens to the power its flat solar panels receive, facing the Sun?

Olympiad problemsPast IOAA problems, restated1 problem

2007How far away is the K star?2 points

Adapted from IOAA 2007 (Chiang Mai), theory problem 1.10. We have restated it in our own words and kept the original data. The solution was last checked on 20 October 2026.

A main-sequence star of type K has a luminosity of $0.4 L_{⊙}$ , and its flux at Earth is $6.23 \times 10^{- 14}$ W m⁻². How far away is it, in parsecs? Ignore the atmosphere, and take $L_{⊙} = 3.828 \times 10^{26}$ W and 1 pc $= 3.086 \times 10^{16}$ m.

I solved it without the solution

AnswerThe star is about 453 pc away.